NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 - Determinants

In this article, you will get NCERT solutions for Class 12 Maths chapter 4 exercise 4.6 consists of questions related to applications of matrix and determinants. There are a lot of applications of matrices and determinants in the field of research, statistics, computer graphics, business, economics, and mathematics. In exercise 4.6 Class 12 Maths you will learn the specific application of matrix and determinant i.e. solving system of linear equation using the inverse of the matrix. All the questions in Class 12 Maths ch 4 ex 4.6 are related to solving the system of linear equations having unique solutions. In this Class 12 Maths chapter 4 exercise 4.6 solutions you will get questions related to finding consistency of linear equations. Also, check NCERT Solutions here.

Also, see

Determinants Exercise 4.1
Determinants Exercise 4.2
Determinants Exercise 4.3
Determinants Exercise 4.4
Determinants Exercise 4.5
Determinants Miscellaneous Exercise

Determinants Exercise:4.6

Question:1 Examine the consistency of the system of equations.

$\small x+2y=2$

$\small 2x+3y=3$

Answer:

We have given the system of equations:18967

$\small x+2y=2$

$\small 2x+3y=3$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &2 \\ 2&3 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 2\\3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(3) -2(2) = -1 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:2 Examine the consistency of the system of equations

$\small 2x-y=5$

$\small x+y=4$

Answer:

We have given the system of equations:

$\small 2x-y=5$

$\small x+y=4$

The given system of equations can be written in the form of matrix; $AX =B$

where $A= \begin{bmatrix} 2 &-1 \\ 1&1 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 2(1) -1(-1) = 3 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:3 Examine the consistency of the system of equations.

$\small x+3y=5$

$\small 2x+6y=8$

Answer:

We have given the system of equations:

$\small x+3y=5$

$\small 2x+6y=8$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &3 \\ 2&6 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\8 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6) -2(3) = 0$

Here A is singular matrix therefore now we will check whether the $(adjA)B$ is zero or non-zero.

$adjA= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}$

So, $(adjA)B= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}\begin{bmatrix} 5\\8 \end{bmatrix} = \begin{bmatrix} 30-24\\-10+8 \end{bmatrix}=\begin{bmatrix} 6\\-2 \end{bmatrix} \neq 0$

As, $(adjA)B \neq 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

Question:4 Examine the consistency of the system of equations.

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

Answer:

We have given the system of equations:

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 1& 1&1 \\ 2& 3& 2\\ a& a &2a \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 1\\2 \\ 4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6a-2a) -1(4a-2a)+1(2a-3a)$

$= 4a -2a-a = 4a -3a =a \neq 0$

[If zero then it won't satisfy the third equation]

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:5 Examine the consistency of the system of equations.

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

Answer:

We have given the system of equations:

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

The given system of equations can be written in the form of matrix; $AX =B$

where $A = \begin{bmatrix} 3& -1&-2 \\ 0& 2& -1\\ 3& -5 &0 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 3(0-5) -(-1)(0+3)-2(0-6)$

$= -15 +3+12 = 0$

Therefore matrix A is a singular matrix.

So, we will then check $(adjA)B,$

$(adjA) = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}$

$\therefore (adjA)B = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}\begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10-10+15\\ -6-6+9 \\ -12-12+18 \end{bmatrix} = \begin{bmatrix} -5\\-3 \\ -6 \end{bmatrix} \neq 0$

As, $(adjA)B$ is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

Question:6 Examine the consistency of the system of equations.

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

Answer:

We have given the system of equations:

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 5& -1&4 \\ 2& 3& 5\\ 5& -2 &6 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 5\\2 \\ -1 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 5(18+10) +1(12-25)+4(-4-15)$

$= 140-13-76 = 51 \neq 0$

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:7 Solve system of linear equations, using matrix method.

$\small 5x+2y=4$

$\small 7x+3y=5$

Answer:

The given system of equations

$\small 5x+2y=4$

$\small 7x+3y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &4 \\ 7& 3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 4\\5 \end{bmatrix}$

we have,

$|A| = 15-14=1 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = (adjA) = \begin{bmatrix} 3 &-2 \\ -7& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 3 &-2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 12-10\\ -28+25 \end{bmatrix} = \begin{bmatrix} 2\\-3 \end{bmatrix}$

Hence the solutions of the given system of equations;

x = 2 and y =-3.

Question:8 Solve system of linear equations, using matrix method.

$2x-y=-2$

$3x+4y=3$

Answer:

The given system of equations

$2x-y=-2$

$3x+4y=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &-1 \\ 3& 4 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} -2\\3 \end{bmatrix}$

we have,

$|A| = 8+3=11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3& 2 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2\\3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -8+3\\ 6+6 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5\\12 \end{bmatrix}= \begin{bmatrix} -\frac{5}{11}\\ \\-\frac{12}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-5}{11} \ and\ y =\frac{12}{11}.$

Question:9 Solve system of linear equations, using matrix method.

$\small 4x-3y=3$

$\small 3x-5y=7$

Answer:

The given system of equations

$\small 4x-3y=3$

$\small 3x-5y=7$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 4 &-3 \\ 3& -5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\7 \end{bmatrix}$

we have,

$|A| = -20+9=-11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{-1}{11}\begin{bmatrix} -5 &3 \\ -3& 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 &-3 \\ 3& -4 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 5 &-3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3\\7 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 15-21\\ 9-28 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -6\\-19 \end{bmatrix}= \begin{bmatrix} -\frac{6}{11}\\ \\-\frac{19}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-6}{11} \ and\ y =\frac{-19}{11}.$

Question:10 Solve system of linear equations, using matrix method.

$\small 5x+2y=3$

$\small 3x+2y=5$

Answer:

The given system of equations

$\small 5x+2y=3$

$\small 3x+2y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &2 \\ 3& 2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\5 \end{bmatrix}$

we have,

$|A| = 10-6=4 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 6-10\\ -9+25 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4\\16 \end{bmatrix}= \begin{bmatrix} -1\\4 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =-1 \ and\ y =4.$

Question:11 Solve system of linear equations, using matrix method.

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

Answer:

The given system of equations

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &1 &1 \\ 1 & -2 &-1 \\ 0& 3 &-5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ and $B =\begin{bmatrix} 1\\ \\ \frac{3}{2} \\ \\ 9 \end{bmatrix}$

we have,

$|A| =2(10+3)-1(-5-0)+1(3-0) = 26+5+3 = 34 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(10+3) = 13$ $A_{12} =(-1)^{1+2}(-5-0) = 5$

$A_{13} =(-1)^{1+3}(3-0) = 3$ $A_{21} =(-1)^{2+1}(-5-3) = 8$

$A_{22} =(-1)^{2+2}(-10-0) = -10$ $A_{23} =(-1)^{2+3}(6-0) = -6$

$A_{31} =(-1)^{3+1}(-1+2) = 1$ $A_{32} =(-1)^{3+2}(-2-1) = 3$

$A_{33} =(-1)^{3+3}(-4-1) = -5$

$(adjA) =\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}\begin{bmatrix} 1\\\frac{3}{2} \\ 9 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 13+12+9\\5-15+27 \\ 3-9-45 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34\\17 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\\frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =\frac{1}{2},\ and\ \ z=-\frac{3}{2}.$

Question:12 Solve system of linear equations, using matrix method.

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

Answer:

The given system of equations

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &1 \\ 2 & 1 &-3 \\ 1& 1 &1 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 4\\ 0 \\ 2 \end{bmatrix}.$

we have,

$|A| =1(1+3)+1(2+3)+1(2-1) = 4+5+1= 10 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(1+3) = 4$ $A_{12} =(-1)^{1+2}(2+3) = -5$

$A_{13} =(-1)^{1+3}(2-1) = 1$ $A_{21} =(-1)^{2+1}(-1-1) = 2$

$A_{22} =(-1)^{2+2}(1-1) = 0$ $A_{23} =(-1)^{2+3}(1+1) = -2$

$A_{31} =(-1)^{3+1}(3-1) = 2$ $A_{32} =(-1)^{3+2}(-3-2) = 5$

$A_{33} =(-1)^{3+3}(1+2) = 3$

$(adjA) =\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}\begin{bmatrix} 4\\0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 16+0+4\\-20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20\\-10 \\ 10 \end{bmatrix}= \begin{bmatrix} 2\\-1 \\ 1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =-1,\ and\ \ z=1.$

Question:13 Solve system of linear equations, using matrix method.

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

Answer:

The given system of equations

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &3 &3 \\ 1 & -2 &1 \\ 3& -1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 5\\ -4 \\ 3 \end{bmatrix}.$

we have,

$|A| =2(4+1) -3(-2-3)+3(-1+6) = 10+15+15 = 40$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(4+1) = 5$ $A_{12} =(-1)^{1+2}(-2-3) = 5$

$A_{13} =(-1)^{1+3}(-1+6) = 5$ $A_{21} =(-1)^{2+1}(-6+3) = 3$

$A_{22} =(-1)^{2+2}(-4-9) = -13$ $A_{23} =(-1)^{2+3}(-2-9) = 11$

$A_{31} =(-1)^{3+1}(3+6) = 9$ $A_{32} =(-1)^{3+2}(2-3) = 1$

$A_{33} =(-1)^{3+3}(-4-3) = -7$

$(adjA) =\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5&11 & -7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}\begin{bmatrix} 5\\-4 \\ 3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25-12+27\\25+52+3 \\ 25-44-21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40\\80 \\ -40 \end{bmatrix}= \begin{bmatrix} 1\\2 \\ -1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=-1.$

Question:14 Solve system of linear equations, using matrix method.

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

Answer:

The given system of equations

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &2 \\ 3 & 4 &-5 \\ 2& -1 &3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 7\\ -5 \\ 12 \end{bmatrix}.$

we have,

$|A| =1(12-5) +1(9+10)+2(-3-8) = 7+19-22 = 4 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{12-5} = 7$ $A_{12} =(-1)^{1+2}(9+10) = -19$

$A_{13} =(-1)^{1+3}(-3-8) = -11$ $A_{21} =(-1)^{2+1}(-3+2) = 1$

$A_{22} =(-1)^{2+2}(3-4) = -1$ $A_{23} =(-1)^{2+3}(-1+2) = -1$

$A_{31} =(-1)^{3+1}(5-8) = -3$ $A_{32} =(-1)^{3+2}(-5-6) = 11$

$A_{33} =(-1)^{3+3}(4+3) = 7$

$(adjA) =\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}\begin{bmatrix} 7\\-5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49-5-36\\-133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8\\4 \\ 12 \end{bmatrix}= \begin{bmatrix} 2\\1 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =1,\ and\ \ z=3.$

Question:15 If $A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , find $A^-^1$ . Using $A^-^1$ solve the system of equations

$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

Answer:

The given system of equations

$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

can be written in the matrix form of AX =B, where

$A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}.$

we have,

$|A| =2(-4+4) +3(-6+4)+5(3-2) = 0-6+5 = -1 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{-4+4} = 0$ $A_{12} =(-1)^{1+2}(-6+4) = 2$

$A_{13} =(-1)^{1+3}(3-2) = 1$ $A_{21} =(-1)^{2+1}(6-5) = -1$

$A_{22} =(-1)^{2+2}(-4-5) = -9$ $A_{23} =(-1)^{2+3}(2+3) = -5$

$A_{31} =(-1)^{3+1}(12-10) = 2$ $A_{32} =(-1)^{3+2}(-8-15) = 23$

$A_{33} =(-1)^{3+3}(4+9) = 13$

$(adjA) =\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = -1\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix} = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}\begin{bmatrix} 11\\-5 \\ -3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6\\-22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1\\2 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=3.$

Question:16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Answer:

So, let us assume the cost of onion, wheat, and rice be x, y and z respectively.

Then we have the equations for the given situation :

$4x+3y+2z = 60$

$2x+4y+6z = 90$

$6x+2y+3y = 70$

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix $A$ .

We have;

$A = \begin{bmatrix} 4 &3 &2 \\ 2& 4 &6 \\ 6 & 2 & 3 \end{bmatrix},$ $X= \begin{bmatrix} x\\y \\ z \end{bmatrix}$ $and\ B = \begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}.$

$|A| = 4(12-12) -3(6-36)+2(4-24) = 0 +90-40 = 50 \neq 0$

Now, we will find the cofactors of A;

$A_{11} = (-1)^{1+1}(12-12) = 0$ $A_{12} = (-1)^{1+2}(6-36) = 30$

$A_{13} = (-1)^{1+3}(4-24) = -20$ $A_{21} = (-1)^{2+1}(9-4) = -5$

$A_{22} = (-1)^{2+2}(12-12) = 0$ $A_{23} = (-1)^{2+3}(8-18) = 10$

$A_{31} = (-1)^{3+1}(18-8) = 10$ $A_{32} = (-1)^{3+2}(24-4) = -20$

$A_{33} = (-1)^{3+3}(16-6) = 10$

Now we have adjA;

$adjA = \begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$ s

So, the solutions can be found by $X = A^{-1}B = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-450+700\\1800+0-1400 \\ -1200+900+700 \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 250\\400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5\\8 \\ 8 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =5,\ y =8,\ and\ \ z=8.$

Therefore, we have the cost of onions is Rs. 5 per Kg, the cost of wheat is Rs. 8 per Kg, and the cost of rice is Rs. 8 per kg.

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

This chapter is very important for board exams and competitive exams like JEE, SRMJEE, etc.
Class 12 Maths chapter 4 exercise 4.6 solutions are helpful for you in solving the system of linear equations and linear equations consistency problems.
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 are helpful for the students who are not able to solve NCERT problems by themself.
In the Class 12th Maths chapter 4 exercise 4.6 solutions, you will learn the art of answering in board exams to get good marks.
You can take these Class 12 Maths chapter 4 exercise 4.6 solutions for reference.

Also see-

NCERT Solutions for Class 12 Maths Chapter 4
NCERT Exemplar Solutions Class 12 Maths Chapter 4

NCERT Solutions of Class 12 Subject Wise

NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

NCERT Exemplar Solutions for Class 12th Maths
NCERT Exemplar Solutions for Class 12th Physics
NCERT Exemplar Solutions for Class 12th Chemistry
NCERT Exemplar Solutions for Class 12th Biology

Happy learning!!!

NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 - Determinants

More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions